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HSC Science (General) ११ वीं कक्षा - Maharashtra State Board Question Bank Solutions for Mathematics and Statistics

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Mathematics and Statistics
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Select the correct answer from the given alternatives.

`lim_(x → π/3) ((tan^2x - 3)/(sec^3x - 8))` =

[2.7] Limits
Chapter: [2.7] Limits
Concept: undefined >> undefined

Select the correct answer from the given alternatives.

`lim_(x -> 0) ((5sinx - xcosx)/(2tanx - 3x^2))` =

[2.7] Limits
Chapter: [2.7] Limits
Concept: undefined >> undefined

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Select the correct answer from the given alternatives.

`lim_(x -> pi/2) [(3cos x + cos 3x)/(2x - pi)^3]` =

[2.7] Limits
Chapter: [2.7] Limits
Concept: undefined >> undefined

Evaluate the following :

`lim_(x -> 0)[(secx^2 - 1)/x^4]`

[2.7] Limits
Chapter: [2.7] Limits
Concept: undefined >> undefined

Evaluate the following :

`lim_(x -> 0) [(x(6^x - 3^x))/(cos (6x) - cos (4x))]`

[2.7] Limits
Chapter: [2.7] Limits
Concept: undefined >> undefined

Evaluate the following :

`lim_(x -> "a") [(sinx - sin"a")/(x - "a")]`

[2.7] Limits
Chapter: [2.7] Limits
Concept: undefined >> undefined

Evaluate the following :

`lim_(x -> "a") [(x cos "a" - "a" cos x)/(x - "a")]`

[2.7] Limits
Chapter: [2.7] Limits
Concept: undefined >> undefined

Evaluate the following :

`lim_(x -> pi/4) [(sinx - cosx)^2/(sqrt(2) - sinx - cosx)]`

[2.7] Limits
Chapter: [2.7] Limits
Concept: undefined >> undefined

Differentiate the following w.r.t.x. :

y = x5 tan x

[2.9] Differentiation
Chapter: [2.9] Differentiation
Concept: undefined >> undefined

Differentiate the following w.r.t.x. :

y = (x2 + 2)2 sin x

[2.9] Differentiation
Chapter: [2.9] Differentiation
Concept: undefined >> undefined

Differentiate the following w.r.t.x. :

y = `"e"^xsecx - x^(5/3) log x`

[2.9] Differentiation
Chapter: [2.9] Differentiation
Concept: undefined >> undefined

Differentiate the following w.r.t.x. :

y = `x^4 + x sqrt(x) cos x - x^2"e"^x`

[2.9] Differentiation
Chapter: [2.9] Differentiation
Concept: undefined >> undefined

Differentiate the following w.r.t.x. :

y = `sinx logx + "e"^x cos x - "e"^x sqrt(x)`

[2.9] Differentiation
Chapter: [2.9] Differentiation
Concept: undefined >> undefined

Differentiate the following w.r.t.x. :

y = `"e"^x tanx + cos x log x - sqrt(x)  5^x`

[2.9] Differentiation
Chapter: [2.9] Differentiation
Concept: undefined >> undefined

Differentiate the following w.r.t.x. :

y = `(x^2 sin x)/(x + cos x)`

[2.9] Differentiation
Chapter: [2.9] Differentiation
Concept: undefined >> undefined

Fill in the blanks:

y = ex .tan x

Differentiating w.r.t.x

`("d"y)/("d"x) = "d"/("d"x)("e"^x tan x)`

= `square "d"/("d"x) tanx + tan x "d"/("d"x) square`

= `square  square + tan x  square`

= `"e"^x [square  + square]`

[2.9] Differentiation
Chapter: [2.9] Differentiation
Concept: undefined >> undefined

Fill in the blanks:

y = `sinx/(x^2 + 2)`

Differentiating. w.r.t.x.

`("d"y)/("d"x) = (square "d"/("d"x) (sin x) - sin x "d"/("dx) square)/(x^2 + 2)^2`

= `(square  square - sin x  square)/(x^2 + 2)^2`

= `(square - square)/(x^2 + 2)^2`

[2.9] Differentiation
Chapter: [2.9] Differentiation
Concept: undefined >> undefined

Fill in the blanks:

y = (3x2 + 5) cos x

Differentiating w.r.t.x

`("d"y)/("d"x) = "d"/("d"x) [(3x^2 + 5) cos x]`

= `(3x^2 + 5) "d"/("d"x) [square] + cos x  "d"/("d"x) [square]`

= `(3x^2 + 5) [square] + cos x  [square]`

∴ `(dx)/("d"y) = (3x^2 + 5) [square] + [square] cos x`

[2.9] Differentiation
Chapter: [2.9] Differentiation
Concept: undefined >> undefined

Fill in the blank:

Differentiate tan x and sec x w.r.t.x. using the formulae for differentiation of `"u"/"v" and 1/"v"` respectively

[2.9] Differentiation
Chapter: [2.9] Differentiation
Concept: undefined >> undefined

Select the correct answer from the given alternative:

If y = `(5sin x - 2)/(4sin x + 3)`, then `("d"y)/("d"x)` =

[2.9] Differentiation
Chapter: [2.9] Differentiation
Concept: undefined >> undefined
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